Instrction Arrangement HDU - 4109 （拓扑排序*）

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最新推荐文章于 2022-11-01 23:18:10 发布

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分类专栏：拓扑排序 vjudge 图论 HDU

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本文探讨了在计算机组织与架构课程中遇到的指令依赖性问题，如WAR、WAW、RAW，并提出了通过调整指令顺序和增加气泡操作来最小化CPU运行时间的解决方案。使用邻接表和拓扑排序算法，确保在无限核心的强CPU上并行执行指令时，避免了因指令间距离不足而产生的危险。

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Instrction Arrangement

HDU - 4109

Ali has taken the Computer Organization and Architecture course this term. He learned that there may be dependence between instructions, like WAR (write after read), WAW, RAW.
If the distance between two instructions is less than the Safe Distance, it will result in hazard, which may cause wrong result. So we need to design special circuit to eliminate hazard. However the most simple way to solve this problem is to add bubbles (useless operation), which means wasting time to ensure that the distance between two instructions is not smaller than the Safe Distance.
The definition of the distance between two instructions is the difference between their beginning times.
Now we have many instructions, and we know the dependent relations and Safe Distances between instructions. We also have a very strong CPU with infinite number of cores, so you can run as many instructions as you want simultaneity, and the CPU is so fast that it just cost 1ns to finish any instruction.
Your job is to rearrange the instructions so that the CPU can finish all the instructions using minimum time.

Input

The input consists several testcases.
The first line has two integers N, M (N <= 1000, M <= 10000), means that there are N instructions and M dependent relations.
The following M lines, each contains three integers X, Y , Z, means the Safe Distance between X and Y is Z, and Y should run after X. The instructions are numbered from 0 to N - 1.

Output

Print one integer, the minimum time the CPU needs to run.

Sample Input

5 2
1 2 1
3 4 1

Sample Output

Hint

In the 1st ns, instruction 0, 1 and 3 are executed;
In the 2nd ns, instruction 2 and 4 are executed.
So the answer should be 2.

#include <bits/stdc++.H>
using namespace std;

queue<int> q;
int head[1111], cnt, indegree[1111], weight[1111];
struct Edge{
	int to, v, w;
}edge[11111];

void addedge(int u, int v, int w) {
	edge[cnt].to = head[u];
	edge[cnt].v = v;
	edge[cnt].w = w;
	head[u] = cnt++;
}

void init() {
	memset(head, -1, sizeof(head)); //邻接表 
	memset(indegree, 0, sizeof(indegree));// 入度 
	memset(weight, 0, sizeof(weight)); //最早开始时间 
	cnt = 0;
}

void tuopupaixu(int n) {
	
	for (int i = 0; i < n; ++i) {
		if (indegree[i] == 0) {
			//printf ("i: %d\n", i);
			q.push(i);
			indegree[i] = -1;
		}
	}
	while (!q.empty()) {
		
		int now = q.front();
		q.pop();
		//printf ("now: %d\n", now);
		for (int i = head[now]; ~i; i = edge[i].to) {
			int v = edge[i].v;
			--indegree[v];
			weight[v] = max(weight[now] + edge[i].w, weight[v]); //更新早开始时间 
			if (indegree[v] == 0)  {
				q.push(v);
				indegree[v] = -1;
			}
		}
		
	}
	/*for (int i = 0; i < n; ++i) {
		printf ("%d ", weight[i]);
	}
	puts("");*/
}

int main() {
	
	int n, m;
	while (~scanf ("%d %d", &n, &m)) {
		init();
		int u, v, w;
		for (int i = 1; i <= m; ++i) {
			scanf ("%d %d %d", &u, &v, &w);
			++indegree[v];
			w = max(w, 1); //每个事件的持续时间是1，所以即使安全时间小于1也要保证能完成事件 
			addedge(u, v, w);
		}
		for (int i = 0; i < n; ++i) {
			if (head[i] == -1) {
				addedge(i, n, 1); // 添加汇点方便求解，每个事件持续时间是1 
			}
		}
		tuopupaixu(n);
		printf ("%d\n", weight[n]); //汇点的最早开始时间=它的最晚开始时间 
	}
	
	return 0;
}